2
$\begingroup$

Find the smallest positive number $p$ for which the equation $\cos(p\sin{x})=\sin(p\cos{x})$ has a solution $x$ belonging $[0,2\pi]$. I am not able to solve this problem. Please help me.

$\endgroup$
2
$\begingroup$

It must be that $p\sin{x}+p\cos{x}=\large\frac{\pi}{2}$ with, $p=\frac{\large\frac{\pi}{2}}{\sin{x}+\cos{x}}$. So, to minimize $p$, $\sin{x}+\cos{x}$ must be maximized.

$\sin{x}+\cos{x}=\sqrt2\sin(x+\pi/4)$, which is maximized when $\sin(x+\pi/4)=1$ at $x=\pi/4$.

Hence, $p=\frac{\large\frac{\pi}{2}}{\sqrt2sin(\pi/2)}$

$p=\frac{\pi}{2\sqrt2}$.

$x=\pi/4$.

$\endgroup$
  • $\begingroup$ the answer in book is pi/2sqrt(2) at x equal to pi/4 and 7pi/4 $\endgroup$ – Jai Mahajan Jul 30 '14 at 10:20
  • $\begingroup$ Please help me in solving this question. How do we know that this answer is the smallest value. $\endgroup$ – Jai Mahajan Jul 30 '14 at 10:21
  • $\begingroup$ I guess this solution is more understandable. $\endgroup$ – Juanito Jul 30 '14 at 19:24
3
$\begingroup$

Given two angles $\alpha$ and $\beta$ one has $$\cos\alpha=\sin\beta=\cos(\beta-{\pi\over2})$$ iff either $$\alpha=\beta-{\pi\over2}+2k\pi\tag{1}$$ or $$\alpha=-(\beta-{\pi\over2})+2k\pi,\quad{\rm i.e.}\quad \alpha+\beta={\pi\over2}+2k\pi\ .\tag{2}$$ In our case $\alpha=p\sin x$ and $\beta=p\cos x$, so that $(1)$ is equivalent with $$p(\cos x-\sin x)={\pi\over2}-2k\pi\ ,$$ so that we have to make sure that the equation $$p{2\over\sqrt{2}}\sin({\pi\over4}-x)={\pi\over2}-2k\pi$$ has a real solution $x$. This is the case if $${2\over\sqrt{2}}p \geq|{\pi\over2}-2k\pi|$$ for a suitable $k$, and the smallest positive $p$ that satisfies this is $p={\pi\over 2\sqrt{2}}$. The resulting solution $x$ of the original equation is then $x={7\pi\over4}$.

The case $(2)$ is similar and has the same outcome for $p$ (so that this is the definitive solution of the problem); the corresponding $x$ is then ${\pi\over4}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.